Question

[Toàn Nguyễn Khánh]

A=\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{100^2}-1\right)\)

So sánh A với \(-\frac{1}{2}\)

Answer 1

[Lê Tài Bảo Châu]

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{100^2}-1\right)\)

\(=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)....\left(1-\frac{1}{100^2}\right)\)

\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}....\frac{100^2-1}{100^2}\)

\(=-\frac{1.3}{2.2}.\frac{2.4}{3.3}....\frac{99.101}{100.100}\)

\(=-\frac{1.2....99}{2.3...100}.\frac{3.4...101}{2.3...100}\)

\(=-\frac{1}{100}.\frac{101}{2}\)

\(=-\frac{101}{200}< \frac{-1}{2}\)

\(\Rightarrow A< \frac{-1}{2}\)

Vậy...